Problem: In Takahashi's mind, there is always an integer sequence of length 2 \times 10^9 + 1: A = (A_{-10^9}, A_{-10^9 + 1}, ..., A_{10^9 - 1}, A_{10^9}) and an integer P.

Initially, all the elements in the sequence A in Takahashi's mind are 0, and the value of the integer P is 0.

When Takahashi eats symbols `+`, `-`, `>` and `<`, the sequence A and the integer P will change as follows:

* When he eats `+`, the value of A_P increases by 1;
* When he eats `-`, the value of A_P decreases by 1;
* When he eats `>`, the value of P increases by 1;
* When he eats `<`, the value of P decreases by 1.



Takahashi has a string S of length N. Each character in S is one of the symbols `+`, `-`, `>` and `<`. He chose a pair of integers (i, j) such that 1 \leq i \leq j \leq N and ate the symbols that are the i-th, (i+1)-th, ..., j-th characters in S, in this order. We heard that, after he finished eating, the sequence A became the same as if he had eaten all the symbols in S from first to last. How many such possible pairs (i, j) are there?

Constraints

* 1 \leq N \leq 250000
* |S| = N
* Each character in S is `+`, `-`, `>` or `<`.

Input

Input is given from Standard Input in the following format:


N
S


Output

Print the answer.

Examples

Input

5
+>+<-


Output

3


Input

5
+>+-<


Output

5


Input

48
-+><<><><><>>>+-<<>->>><<><<-+<>><+<<>+><-+->><<


Output

475

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
int n,i;
char c[500005];
unsigned long long h,s[250005],ans,t;
map<unsigned long long,unsigned long long> m;
unsigned long long qpow(unsigned long long a,unsigned long long b)
{
	unsigned long long ans=1;
	while(b)
	{
		if(b&1)
			ans=ans*a;
		a=a*a;
		b>>=1;
	}
	return ans;
}
int main()
{
	scanf("%d",&n);
	scanf("%s",c+1);
	unsigned long long inv=qpow(250001,9223372036854775807ull);
	s[0]=1;
	for(i=1;i<=n;++i)
	{
		s[i]=s[i-1];
		if(c[i]=='<')
			s[i]*=inv;
		if(c[i]=='>')
			s[i]*=250001;
	}
	for(i=n;i>=1;--i)
	{
		if(c[i]=='+')
			++t;
		if(c[i]=='-')
			--t;
		if(c[i]=='<')
			t=t*inv;
		if(c[i]=='>')
			t=t*250001;
	}
	++m[0];
	for(i=n;i>=1;--i)
	{
		if(c[i]=='+')
			++h;
		if(c[i]=='-')
			--h;
		if(c[i]=='<')
			h=h*inv;
		if(c[i]=='>')
			h=h*250001;
		ans+=m[(h-t)*s[i-1]];
		++m[h*s[i-1]];
	}
	cout<<ans;
}